Puzzles

SAM is 781.

Because 13 and 11 are prime and larger than 7, no-one can have taken them, as their is no way the other person's product could then be the same. Once 13 and 11 are discarded, the prime factorisation of the product of all the other cards is $2^{11}\times 3^5\times 5^2\times 7^2$. In order to be shared to give the same product, the powers must be even, and so the other card not taken must be 6 (as 2 and 3 are the numbers with odd powers).

Therefore the product of the three cards is 858.

30030 is 2×3×5×7×11×13, so the answer is the number of ways of sharing six elements into unlabelled boxes (A000110).

The answer is 203.

For any digit $V$, $VVV$ is a multiple of 111, and $111=37\times 3$. 37 is prime, so $VI$ must be a multiple of 37 (as $X$ is less than 10 so cannot be a multiple of 37). Therefore $VI$ is either 37 or 74.

If $VI$ was 74, then $VVV$ is 777. But then $VI\times X$ is even and $VVV$ is odd. This is impossible, so $VI$ must the 37.

Now that we know that $VI$ is 37, we see that

and so $X=9$ and the final solution is

EDCCBA is a multiple of four, so A is even. A cannot be more than 2, as otherwise EDCCBA would have more digits. So A is 2.

E must therefore be 8 or 9 (as 4 times B is less than E) and 3 or 8 (as 4 times E ends in 2). Therefore E is 8.

Carrying on like this, we find:

The number of 0s at the end of 1000! will be equal to the number of times 10 is a factor of it. For each 10 that is a factor, there will be a 5 and a 2 that are factors. Therfore the number of 0s will be equal to the number of times 5 is a factors of 100! (because 5 is larger than 2).

Therefore the number of 0s at the end of 1000! will be equal to $\lfloor \frac{1000}{5}\rfloor +\lfloor \frac{1000}{25}\rfloor +\lfloor \frac{1000}{125}\rfloor +\lfloor \frac{1000}{625}\rfloor$. This is 249.

205

The number of zeros at the end of a number is the same as the number of 10s in the product that makes the number. Each of these 10s is made by multiplying 5 by 2.

There will be more even numbers than multiples of 5 in $100!$, so the number of 5s will tell us how many zeros the number ends in.

In $100!$, there will be 20 multiples of 5 and 4 multiples of $5^2$. This means that $100!$ will end in 24 zeros.

How many zeros will $n!$ end in?